UPSC NDA I 2024 Mathematics Paper

Concept:This problem involves summing an arithmetico-geometric series and using the formula for the sum of a geometric progression (GP).Explanation:The given data is:Thus, $\sum_i^n x_i f_i = \frac{1}{1} + \frac{2}{2} + \frac{3}{2^2} + \cdots + \frac{n}{2^{n-1}}$.Let $S = \frac{1}{1} + \frac{2}{2} + \frac{3}{2^2} + \cdots + \frac{n}{2^{n-1}}$.Multiplying both sides by $\frac12$ gives $\frac12 S = \frac{1}{2} + \frac{2}{2^2} + \frac{3}{2^3} + \cdots + \frac{n}{2^{n}}$.Subtract the second equation from the first: $S - \frac12 S = \frac12 S = \left(1 + \frac12 + \frac1{2^2} + \cdots + \frac1{2^{n-1}}\right) - \frac{n}{2^n}$.The series in parentheses is a GP with first term $1$, common ratio $\frac12$, and $n$ terms. Its sum is $\frac{1(1-(\frac12)^n)}{1-\frac12} = 2\left(1-\frac1{2^n}\right)$.So $\frac12 S = 2\left(1-\frac1{2^n}\right) - \frac{n}{2^n} = 2 - \frac{2}{2^n} - \frac{n}{2^n} = 2 - \frac{n+2}{2^n}$.Multiply by $2$: $S = 4 - \frac{n+2}{2^{n-1}} = \frac{4\cdot 2^{n-1} - (n+2)}{2^{n-1}} = \frac{2^{n+1} - n - 2}{2^{n-1}}$.Thus, $\sum_i^n x_i f_i = \frac{2^{n+1} - n - 2}{2^{n-1}}$.Answer:$\frac{2^{n+1} - n - 2}{2^{n-1}}$